Method of undetermined coefficients for y''+y'+y=cos(x)-x²eˣ

[ProPatto16]

solve y''+y'+y=cosx-x²e^x

y= y_c+y_p

y_c:

characteristic eq gives r²+r+1=0

using quadratic formula i got r=-1/2+[(sqrt3)/2]i and r=-1/2-[(sqrt3)/2]i

so y_c=e^-x/2(c₁cos(sqrt3/2)x+c₂sin(sqrt3/2)x

but i have no idea of a particular solution that makes any sense.

i tried a general approach, the functions and their derivitives on the RHS include terms cosx, sinx, x²e^x, xe^x, e^x

so i good start would be y_p=Acosx+Bsinx+Cx²e^x+Dxe^x+Ee^x

then find y'_p and y''_p and sub into original equation.
but it becomes so dam monstrous there must be another way.

 

[vela]

You just have to grind through it. You might find it easier to deal with cos x and x²e^x separately and then add the two particular solutions together at the end.

When dealing with the non-trig terms, factor the common exponential out so that you have a polynomial multiplied by the exponential. That way, when you differentiate, you only have to apply the product rule once.

 

[ProPatto16]

so i grinded through it and ended up with

y_p=0cosx+1sinx-(1/3)x²e^x+(2/3)xe^x-(4/9)e^x

then to solve i put y=y_c+y_p

so y= e^-2/x[(c₁cos(sqrt3/2)x+c₂sin(sqrt3/2)x] + 0cosx+1sinx-(1/3)x²e^x+(2/3)xe^x-(4/9)e^x

yeah?

that seems mighty ridiculous :(

and what about c₁ and c₂?? i don't have initial conditions given so i can't solve for them??

 

[vela]

I'm guessing it's just a typo, but the first exponential should be e^-x/2, not exp^-2/x.

You're correct. You need initial conditions to find the arbitrary constants.